3.120 \(\int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)
Optimal. Leaf size=91 \[ -\frac {(a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a f} \]
[Out]
-1/2*(a-b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(3/2)/f-1/2*cot(f*x+e)*csc(f*x+e)*(a-b+b*s
ec(f*x+e)^2)^(1/2)/a/f
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Rubi [A] time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{3664, 471, 12, 377, 207} \[ -\frac {(a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
-((a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*a^(3/2)*f) - (Cot[e + f*x]*Csc[e
+ f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*a*f)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 471
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 3664
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rubi steps
\begin {align*} \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f}+\frac {\operatorname {Subst}\left (\int \frac {a-b}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f}+\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f}+\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{-1+a x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a f}\\ &=-\frac {(a-b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f}\\ \end {align*}
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Mathematica [B] time = 3.20, size = 303, normalized size = 3.33 \[ -\frac {\cot (e+f x) \csc (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt {2} \sqrt {a} \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}+4 (a-b) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {a-(a-2 b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+4 (a-b) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )\right )}{4 a^{3/2} f \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
-1/4*(Cot[e + f*x]*Csc[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(Sqrt[2]*Sqrt[a]*Sqrt[
(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4] + 4*(a - b)*ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/
(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]*Sin[(e + f*x)/2]^2 + 4*(a - b)*ArcTanh
[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]*S
in[(e + f*x)/2]^2))/(a^(3/2)*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])
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fricas [A] time = 0.63, size = 284, normalized size = 3.12 \[ \left [\frac {2 \, a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, \frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/4*(2*a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - ((a - b)*cos(f*x + e)^2 - a + b)*sq
rt(a)*log(-2*((a - b)*cos(f*x + e)^2 + 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e
) + a + b)/(cos(f*x + e)^2 - 1)))/(a^2*f*cos(f*x + e)^2 - a^2*f), 1/2*(((a - b)*cos(f*x + e)^2 - a + b)*sqrt(-
a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + a*sqrt(((a - b)*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch
eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable
to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)
Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nos
tep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi
/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>
(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost
ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/
t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign:
(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check
sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to
check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab
le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*
pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to chec
k sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable t
o check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Un
able to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_noste
p/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t
_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-
2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep
/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_
nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2
*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check si
gn: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch
eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable
to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)
Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nos
tep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi
/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>
(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost
ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs
or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Disconti
nuities at zeroes of t_nostep^2-1 were not checkedWarning, integration of abs or sign assumes constant sign by
intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Warning, need to choose a branch for the
root of a polynomial with parameters. This might be wrong.The choice was done assuming [a,b]=[85,53]Warning,
need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done
assuming [a,b]=[-33,71]Discontinuities at zeroes of t_nostep^2-1 were not checkedUnable to check sign: (4*pi/t
_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (
2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-
1)]Evaluation time: 3.42Unable to divide, perhaps due to rounding error%%%{1,[4,0]%%%}+%%%{%%%{-2,[1]%%%},[2,0
]%%%}+%%%{%%%{1,[2]%%%},[0,0]%%%} / %%%{%%%{1,[1]%%%},[4,0]%%%}+%%%{%%%{-2,[2]%%%},[2,0]%%%}+%%%{%%%{1,[3]%%%}
,[0,0]%%%} Error: Bad Argument Value
________________________________________________________________________________________
maple [B] time = 1.20, size = 2801, normalized size = 30.78 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x)
[Out]
1/4/f*(((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((
a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^
(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*cos(f*x+e)^3*a^2-((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1
+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*
x+e)+b)/sin(f*x+e)^2/a^(1/2))*cos(f*x+e)^3*a*b+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-
4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)^3*a^2-((a*cos(f
*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)
/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-b*cos
(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)^3*a*b-2*cos(f*x+e)^2*a^(5/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(
f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2
)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*
x+e)^2/a^(1/2))*cos(f*x+e)^2*a^2-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+
e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*cos(f*x+e)^2*a*b+((
a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e
)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)^2*a^2-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)
*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x
+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)^2*a*b+2*co
s(f*x+e)^2*a^(3/2)*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)
*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+c
os(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*cos(f*x+e)*a^2+((a*cos(f*x+e)^2
-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x
+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*cos(f*x+e)*a*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^
(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-co
s(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)*a^2+(
(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e
)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+
e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*cos(f*x+e)*a*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*
ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f
*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*a
^2+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*co
s(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2
)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*a*b-((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x
+e))^2)^(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x
+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*a^2+((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+a*cos(f*x+e)-
b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*a*b-2*a^(3/2)*b)*sin(f*x+e)^2/(-1+cos(f*x+e))^2/cos(f*x+e)/((a*cos(f*x+e)^2-c
os(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/(1+cos(f*x+e))^2/a^(5/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(csc(f*x + e)^3/sqrt(b*tan(f*x + e)^2 + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2)),x)
[Out]
int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**3/(a+b*tan(f*x+e)**2)**(1/2),x)
[Out]
Integral(csc(e + f*x)**3/sqrt(a + b*tan(e + f*x)**2), x)
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